CHAPTER XVI. HEAD RESISTANCE CALCULATIONS.

CHAPTER XVI. HEAD RESISTANCE CALCULATIONS.Effect of Resistance. Resistance to the forward motion of an aeroplane can be divided into two classes, (1) The resistance or drag due to the lift of the wings, and (2) The useless or "Parasitic" resistance due to the body, chassis and other structural parts of the machine. The total resistance is the sum of the wing drag and the parasitic resistance. Since every pound of resistance calls for a definite amount of power, it is of the greatest importance to reduce this loss to the lowest possible amount. The adoption of an efficient wing section means little if there is a high resistance body and a tangle of useless struts and wires exposed to the air stream. The resistance has a much greater effect on the power than the weight.Weight and Resistance. We have seen that the average modern wing section will lift about 16 times the value of the horizontal drag, that is, an addition of 16 pounds will be equal to 1 pound of head resistance. If, by unnecessary resistance, we should increase the drag by 10 pounds, we might as well gain the benefit of 10 x 16 = 160 pounds of useful load. The higher the lift-drag efficiency of the wing, the greater will be the proportional loss by parasitic resistance.Gliding Angle. The gliding angle, or the inclination of the path of descent when the machine is operating without power, is determined by the weight and the total head resistance. With a constant weight the angle is greatest when the resistance is highest. Aside from considerations of power, the gliding angle is of the greatest importance from the standpoint of safety. The less the resistance, and the flatter the angle of descent, the greater the landing radius.Numerically this angle can be expressed by: Glide = W/R, where W = the weight of the aeroplane, and R = total resistance. Thus if the weight is 2500 pounds and the head resistance is 500 pounds, the rate of glide will be: 2500/500 = 5. This means that the machine will travel forward 5 feet for every foot that it falls vertically. If the resistance could be decreased to 100 pounds, the rate of glide would be extended to 2500/100 = 25, or the aeroplane would travel 25 feet horizontally for every foot of descent. This will give an idea as to the value of low resistance.Resistance and Speed. The parasitic resistance of a body in uniform air varies as the square of the velocity at ordinary flight speeds. Comparing speeds of 40 and 100 miles per hour, the ratio will be as 40° is to 100° = 1600: 10,000 = 6.25, that is, the resistance at 100 miles per hour will be 6.25 times as great as at 40 miles per hour.The above remarks apply only to bodies making constant angle with the air stream. Wings and lifting surfaces make varying angles at different speeds and hence do not show the same rate of increase. In carrying a constant load, the angle of the aeroplane wing is decreased as the speed increases and up to a certain point the resistance actually decreases with an increase in the speed. The wing resistance is greatest at extremely low speeds and at very high speeds. As the total resistance is made up of the sum of the wing and parasitic resistance at the different speeds, it does not vary according to any fixed law. The only true knowledge of the conditions existing through the range of flight speeds is obtained by drawing a curve in which the sums of the drag and head resistance are taken at intervals.Resistance and Power. The power consumed in overcoming parasitic resistance increases at a higher rate than the resistance, or as the cube of the speed. Thus if the speed is increased from 40 to 100 miles per hour, the power will be increased 15.63 times. This can be shown by the following: Let V = velocity in miles per hour, H = Horsepower, K= Resistance coefficient of a body, A = Total area of presentation, and R = resistance in pounds. Then H = RV/375. Since R = KAV², then H = KAV² x V/375 = KAV³/375.Resistance and Altitude. The resistance decreases with a reduction in the density of the air at constant speed. In practice, the resistance of an aeroplane is not in direct proportion to a decrease in the density as the speed must be increased at high altitudes in order to obtain the lift. The following example given by Capt. Green will show the actual relations.Taking an altitude of 10,000 feet above sea level where the density is 0.74 of that at sea level, the resistance at equal speeds will be practically in proportion to the densities. In order to gain sustentation at the higher altitude, the speed must be increased, and hence the true resistance will be far from that calculated by the relative densities. Assume a sea level speed of 100 ft./sec., a weight of 3000 pounds, a lift-drag ratio of L/D = 15, and a body resistance of 40 pounds at sea level.Because of the change in density at 10,000 feet, the flying speed will be increased from 100 feet per second to 350 feet per second in order to obtain sustentation. With sea level density this increase in speed (3.5 times) would increase the body resistance 3.5 x 3.5 = 12.25 times, making the total resistance 12.25 x 40 = 490 pounds. Since the density at the higher altitude is only 0.74 of that at sea level, this will be reduced by 0.26, or 0.26 x 490 = 364 pounds. Thus, the final practical result is that the sea level resistance of the body (40 pounds) is increased 9.1 times because of the speed increase necessary for sustentation. Since the wing angle and hence the liftdrag ratio would remain constant under both conditions, the wing drag would be constant at both altitudes, or 3000/15 = 200 pounds. The total sea level resistance at 100 feet per second is 200 + 40 = 240 pounds, while the total resistance at 10,000 feet becomes 364 + 200 = 564 pounds.The speed varies as the square root of the change in density percentage. If V = velocity at sea level, v = velocity at a higher level, and d = percentage of the sea level density at the higher altitude, then v = V/√D. When the velocity at the high altitude is thus determined, the resistance can be easily obtained by the method given in Capt. Green's article. The following table gives the percentage of densities referred to sea level density.Altitude FeetDensity PercentAltitude FeetDensity PercentSea-level1.007,5000.781,000.9710,000.742,000.9512,500.663,000.9115,000.615,000.8520,000.52If the velocity at sea level is 100 miles per hour, the velocity at 20,000 feet will be 100/0.72 = 139 miles per hour, where 0.72 is the square root of the density percentage, or the square root of 0.52 = .72 at 20,000 feet.Total Parasitic Resistance. Aside from the drag of the wings, the resistance of the structural parts, body, tail and chassis depends upon the size and type of aeroplane. A speed scout has less resistance than a larger machine because of the small amount of exposed bracing, although the relative resistance of the body is much greater. The type of engine also has a great influence on the parasitic resistance. The following gives the approximate distribution of a modern fighting aeroplane:Body62 percentLanding gear16 "Tail, fin, rudder7 "Struts, wires, etc.15 "The body resistance is by far the greatest item. A great part of the body resistance can be attributed to the motor cooling system, since in either case it is diverted from the true streamline form in order to accommodate the radiator, or the rotary motor cowl. The body resistance is also influenced by the necessity of accommodating a given cargo or passenger-carrying capacity, and by the distance of the tail surfaces from the wings. A body is not a streamline form when its length greatly exceeds 6 diameters.Calculation of Total Resistance. The nearest approach that we can make to the actual head resistance by means of a formula is to adopt an expression in the form of R = KV² where K is a factor depending upon the size and type of machine. The true method would be to go over the planes and sum up the individual resistance of all the exposed parts. The parts lying in the propeller slip stream should be increased by the increased velocity of the slip stream. The parasitic resistance of biplanes weighing about 1800 pounds will average about, R = 0.036V² where V = velocity in miles per hour. Biplanes averaging 2500 pounds give R = 0.048V². Machines of the training or 2-seater type weigh from 1800 to 2500 pounds, and have an average head resistance distribution as follows:Body, radiators, shields35.5 percent.Tail surface and bracing14.9 "Landing gear17.2 "Interplane struts, wires and fittings23.6 "Ailerons, aileron bracing, etc.8.8 "The averages in the above table differ greatly from the values given for the high speed fighting machine, principally because of the large control surfaces used in training machines, and the difference in the size of the motors.With the wing drag being equal to D = Kx AV², and the total parasitic resistance equal to R = KV², the total resistance can be expressed by Rt = KxAV² + KV², where K = coefficient of parasitic resistance for different types and sizes of machines. The value of K for training machines will average 0.036, for machines weighing about 2500 pounds K = 0.048. Scouts and small machines will be safe at K = 0.028. The wing drag coefficient Kx varies with the angle of incidence and hence with the speed. For example, we will assume that the wing drag (Kx) of a scout biplane at 100 miles per hour is 0.00015, that the area is 200 square feet, and that the parasitic resistance coefficient is K = 0.028. The total resistance becomes: R = (0.00015 x 200 x 100 x 100) + 0.028 x 100 x 100 = 300 + 280 = 580 pounds. The formula in this case would be R = KxAV² + 0.028V².Strut Resistance. The struts are of as nearly streamline form as possible. In practice the resistance must be compromised with strength, and for this reason the struts having the least resistance are not always applicable to the practical aeroplane. From the best results published by the N. P. L. the resistance was about 12.8 pounds per 100 feet strut at 60 miles per hour. The width of the strut is 1 inch. A rectangular strut under the same conditions gave a resistance of 104.4 pounds per 100 feet. A safe value would be 25 pounds per 100 feet at 60 miles per hour. If a wider strut is used, the resistance must be increased in proportion. With a greater speed, the resistance must be increased in proportion to the squares of the velocity. When the struts are inclined with the wind, the resistance is much decreased, and this is one advantage of a heavy stagger in a biplane.The "Fineness ratio" or the ratio of the width to the depth of the section has a great effect on the resistance. With the depth equal to twice the width measured across the stream, a certain strut section gave a resistance of 24.8 pounds per 100 feet, while with a ratio of 3.5 the resistance was reduced to 11.4 pounds per hundred feet. Beyond this ratio the change is not as great, for with a ratio of 4.6 the resistance only dropped to 11.2 pounds.Radiator Resistance. For the exact calculation of the radiator resistance it is first necessary to know the motor power and the fuel consumption since the radiator area, and hence the resistance, depends upon the size of the motor and the amount of heat transmitted to the jacket water. An aeronautic motor may be considered to lose as much through the water jackets as is developed in useful power, so that on this basis we should allow about 1.6 square feet of radiation surface per horsepower. This figure is arrived at by J. C. Hunsaker and assumes that the wind speed is 50 miles per hour (73 feet per second). The most severe cooling condition is met with in climbing at low speed, and it is here assumed that 50 miles per hour will represent the lowest speed that would be maintained for any length of time with the motor full out. For a racing aeroplane that will not climb for any length of time, one-half of the surface given above will be sufficient, and if the radiator is placed in the propeller slip stream it can be made relatively still smaller as the increased propeller slip at rapid rates of climb partially offsets the additional heating.In the above calculations, Hunsaker does not take any particular type of radiator into consideration, merely assuming a smooth cooling surface. The Rome–Turney Company states that they allow 1.08 square feet of cooling surface per horsepower for honeycomb radiators, and 0.85 square feet for the helical tube type. The surface referred to means the actual surface measured all over the tubes and cells, and does not refer to the front area nor the exterior dimensions of the radiator. While a radiator may be made 25 percent smaller when placed in the slipstream, the resistance is increased by about 25 per cent, with a very small saving in weight, hence the total saving is small, if any. Side mounted radiators have a lower cooling effect per square foot than those placed in any other position, owing to the fact that the air must pass through a greater length of tube than where the broad side faces the wind.In the radiator section tested by Hunsaker, there were about 64 square feet of cooling surface per square foot of front face area, but for absolute assurance on this point one should determine the ratio for the particular type of radiator that is to be used. The Auto Radiator Manufacturing Corporation, makers of the "Flexo" copper core radiators, have published some field tests made under practical conditions and for different types and methods of mounting. The four classes of radiators described are: (1) Front Type, in which the radiator is mounted in the end of the fuselage; (2) Side Type, mounted on the sides of the body; (3) Overhead Type, mounted above the fuselage and near the top plane; (4) Over-Engine Type, placed above and connected directly to the motor, as in the Standard H-3.The following table gives the effectiveness of the different mountings in terms of the frontal area required per horsepower and the cooling surface, the area being in square inches (Front face area of radiator). Area in wind of type (3) is half the calculated frontal area since one core lies behind the other: Taking the value of the Rome–Turney honeycomb radiator as 60 square feet of cooling surface per horsepower, the frontal area per horsepower will be 0.0169 square feet, assuming that the radiator is approximately 6 inches thick. This amounts to 2.43 square inches of frontal area per horsepower.Example. Find the approximate frontal area of a Rome-Turney type honeycomb radiator used with a motor giving 100 brake-horsepower. Find Resistance at 50 miles per hour (73 feet per second).Class of MountingSquare Inches Per H.P.Cooling Surface Per H.P Square InchFront Type4.00117.00Side Type7.20104.00Overhead Type2.70112.00Over-Engine Type5.00121.00Solution. Area = A = 0.0169 HP = 0.0169 x 100 = 1.69 square feet. The honeycomb portion of surface for a square radiator of the above area will measure 16.2" x 16.2". Allowing a 1-inch water passage or frame all around the core, the side of the completed square radiator will measure 16.2" + 1" = 18.2". The diameter of a circular radiator core of the same 1.69 x 144 area will be 17.4 inches, since D = 1.69x144/0.7854. Adding the water passage, the overall diameter becomes 17.4 + 1 + 1 = 19.4 inches. The round honeycomb front radiator used on the 100 horsepower Curtiss Baby Scout measures 20 inches. Hunaker found the resistance of a honeycomb radiator to be R=0.000814 AV², there being 4 honeycomb cells per square inch. A = area of radiator in square feet, and V = velocity in feet per second. Adopting, for example, a speed of 73 feet per second, and an area equivalent to a 19.4-inch diameter circular radiator as above, the total resistance becomes:R = 0.000814 AV² = 0.000814 x 3.1x (73 x 73) = 13.32 pounds, at 50 M. P. H. where A = 3.1 square feet.Resistance of Chassis. Disc wheels (Enclosed spokes) have a resistance of about one-half that of open-wire wheels. The N. P. L. and Eiffel have agreed that the resistance of a wheel approximating 26" x 4" has a resistance of 1.7 pounds at 60 miles per hour (Disc type). For any other speed, the wheel resistance will be R = 1.7 V²/3600, where V = speed in miles per hour. We must also take into consideration the axle, chassis struts, wiring, shock absorbers, etc. The itemization of the chassis resistance, as given by the N.P.L. for the B.E.-2 biplane is as follows (60 miles per hour):Wheels 2(a)1.75 pounds3.5 poundsAxle2.0 "Chassis struts and connections1.1 "Total chassis resistance(3)60 MPH.6.6 poundsAt any other speed, the resistance for the complete chassis can be given by the formula R = 6.6V²/3600. This allowance will be ample, as the B. E.-2 is an old type and is equipped with skids.Interplane Resistance. The interplane struts and wires are difficult to estimate by an approximate formula, the only exact way being to figure up each item separately from a preliminary drawing. The resistance varies with the form of the strut or wire section, the length, and the thickness. The fact that some of the struts lie in the propeller slipstream, and some outside of it, makes the calculation doubly difficult. The only recourse that we have at present is to analyze the conditions on the B. E.-2. With struts approximating true streamline form, a great percentage of the total resistance is skin friction, and as before explained, this item varies at a lesser rate than the square of the speed.INTERPLANE RESISTANCE OF BIPLANE B.E-2 AT 60 M.P.H TableAccording to a number of experiments on full size biplanes averaging 1900 pounds, it has been found that the interplane resistance (Struts, wires and fittings) amounts to about 24 per cent of the total parasitic head resistance of the entire machine, the drag of wings not being included. The maximum observed gave 29 per cent and the minimum 15 per cent. The resistance of the interplane bracing of speed scouts will be considerably less in proportion, as there are fewer exposed struts and cables on this type, the resistance probably averaging 15 per cent of the total head resistance. Based on these figures the resistance of the interplane bracing can be expressed by the following formula, in which I = resistance of interplane bracing in pounds, and V = translational speed in miles per hour:I = 0.009 V² (For two-place biplanes weighing 1900 pounds).I = 0.0054V² (For biplane speed scouts or racing type biplanes).Strut Resistance. The above estimate includes wiring, strut fittings, etc., complete, and also takes the effect of the slipstream into consideration. A more accurate estimate can be made on the basis of strut length. To obtain this unit value we have recourse to the B. E.-2 tests. The translational speed in 60 miles per hour (88 feet per second) and the slipstream is taken at 25 feet per second. This gives a total velocity in the slipstream of 113 feet per second. The struts are 1% inches wide, and vary in length from 3’ 0" to 6’ 0". In the slipstream the increased velocity increases the resistance of the items by 64 per cent.Total running length = 110’–0". Total resistance = 10.81 pounds. The resistance per foot = 10.81/110 = 0.099 pounds.Resistance of Wire and Cable. In this estimate we will take the resistance given in the B. E.-2 tests, since values are given in the slipstream as well as for the outer portions. In the translational stream there is 240’ 0" of cable, 70’ 0" of No. 12 solid wire, and 52 turnbuckles, the total giving a resistance of 38.10 pounds. In the slipstream there is 50’ 0" of cable and 30’ 0" of solid wire with a resistance of 11.00 pounds. The total wire and cable resistance for the wings is therefore 49.10 pounds. The resistance of the wire and cable combined is 0.127 pounds per running foot.Summary of Interplane Resistance. The total interplane resistance includes the struts, wires, cables and turnbuckles, a portion of which are in the slipstream. Since the total head resistance of the entire machine (B.E.-2) is 140 pounds at 60 M. P. H., and the interplane resistance = 10.81 + 49.10= 59.91 pounds, the relation of the interplane resistance to the total resistance is 43 per cent. This is much higher than the average (24 per cent), but the B.E.-2 is an old type of machine and the number of struts and wires were much greater than with modern aeroplanes.Control Surface Resistance. The resistance of the control surfaces is a variable quantity, since so much depends upon the arrangement and form. Another variation occurring among machines of the same make and type is due to the various angles of the surfaces during flight, or at least during the time that they are used in correcting the attitude of the machine. With the elevator flaps or ailerons depressed to their fullest extent, the drag is many times that with the surfaces in "neutral," and as a general thing the controls are depressed at the time when the power demand is the greatest—that is, on landing, flying slow, or in "getting off."Ailerons "in neutral" can be considered as being an integral part of the wings when they are hinged to the wing spar. In the older types of Curtiss machines the ailerons were hinged midway between the planes and the resistance was always in existence, whether the ailerons were in neutral or not. Wing warping, in general, can be assumed as in the case where the wings and ailerons are combined. With ailerons built into the wings, the resistance of the ailerons, and their wires and fittings, can be taken as being about 4 per cent of the total head resistance. With the aileron located between the two wings, the resistance may run as high as 20 per cent of the total.Like the ailerons, the elevator surfaces and rudder are variable in attitude and therefore give a varying resistance. In neutral attitude the complete tail, consisting of the rudder, stabilizer, elevator, fin and bracing, will average about 15 per cent of the total resistance, it being understood that a non-lifting stabilizer is fitted. With lifting tails the resistance will be increased in proportion to the load carried by the stabilizer. In regard to the tail resistance it should be noted that these surfaces are in the slipstream and are calculated accordingly, although the velocity of the slipstream is somewhat reduced at the point where it encounters the tail surfaces. The total tail resistance of the B. E.-2 is given as 3.3 pounds.Resistance of Seaplane Floats. The usual type of seaplane with double floats may be considered as having about 12 per cent higher resistance than a similar land machine. Some forms of floats have less resistance than others, owing to their better streamline form, but the above figure will be on the safe side for the average pontoon. Basing our formula on a 12 per cent increase on the total head resistance, the formula for the floats and bracing will become: Rt = 0.00436V² where R1 = resistance of floats and fittings.Body Resistance. This item is probably the most difficult of any to compute, owing to the great variety of forms, the difference in the engine mounting, and the disposition of the fittings and connections. The resistance of the pilot's and passenger's heads, wind shields, and propeller arrangement all tend to increase the difficulty of obtaining a correct value. Aeroplanes with rotary air-cooled motors, or with large front radiators have a higher resistance than those arranged with other types of motors or radiator arrangements. Probably the item having the greatest influence on the resistance of the fuselage is the ratio of the length to the depth, or the "fineness ratio." In tractor monoplanes and biplanes, of the single propeller type, the body is in the slipstream, and compensation must be made for this factor.If it were not for the motor and radiator, the tractor fuselage could be made in true dirigible streamline form, and would therefore present less resistance than the present forms of "practical" bodies. The necessity of placing the tail surfaces at a fixed distance from the wings also involves the use of a body that is longer in proportion than a true streamline form, and this factor alone introduces an excessive head resistance. The ideal ratio of depth to length would seem to range from 1 to 5.5 or 1 to 6. The fineness ratio of the average two-seat tractor is considerably greater than this, ranging from 1 to 7.5 or 8.5. A single-seat machine of the speed-scout type can be made much shorter and has more nearly the ideal proportions.The only possible way of disposing of this problem is to compare the results of wind tunnel tests made on different types of bodies, and even with this data at hand a liberal allowance should be made because of the influence of the connections and other accessories. Eiffel, the N. P. L., and the Massachusetts Institute of Technology have made a number of experiments with scale models of existing aeroplane bodies. It is from these tests that we must estimate our body resistance, hence a table of the results is attached, the approximate outlines being shown by the figures.As in calculating the resistance of other parts, the resistance of the body can be expressed by R = KxAV², where Kx = coefficient of the body form, A = Cross-sectional area of body in square feet (Area of presentation), and V = velocity in miles per hour. The area A is obtained by multiplying the body depth by the width. The "area of presentation" of a body 2’ 6" wide and 3’ 0" deep will be 2.5 x 3 = 7.5 square feet.The experimental data does not give a very ready comparison between the different types, as the bodies not only vary in shape and size, but are also shown with different equipment. Some have tail planes and some have not; two are shown with the heads of the pilot and passenger projecting above the fuselage, while the remainder have either a simple cock-pit opening or are entirely closed. The presence of the propeller in two cases may have a great deal to do with raising the value of the experimental results. The propeller was stationary during the tests, but it was noted that the resistance was considerably less when the propeller was allowed to run as a windmill, driving the motor. This latter condition would correspond to the resistance in gliding with the motor cut off. In all cases, except the Deperdussin, the bodies are covered with fabric, and the sagging of the cloth in flight will probably result in higher resistance than would be indicated by the solid wood or metal model used in the tests. The pusher type bodies give less resistance than the tractors, but the additional resistance of the outriggers and tail bracing will probably bring the total far above the tractor body.In the accompanying body chart are shown 7 representative bodies: (a) Deperdussin Monocoque Monoplane Body, a single-seater; (b) N. P. L.-5 Tractor Biplane Body, single-seater; (c) B. F.-36 Dirigible Form, without propeller or cock-pit openings; (d) B. E.-3. Two-Place Tractor Body, with passenger and pilot; (e) Curtiss JN Type Tractor Body, with passengers, chassis and tail; (f) Farman Pusher type, with motor, propeller and exposed passengers; (g) N. P. L. Pusher Body, bare. Body (a) was tested with a 1/5 scale model at a wind tunnel speed of 28 meters per second, the resistance of the model being 0.377 kilograms (0.83 pounds). Body (d) in model form was 1/16 scale and was tested at 20.5 miles per hour, at which speed the resistance was 0.0165 pounds. Model (e) was 1/12 scale and was tested at 30 miles per hour. These varying test speeds, it will be seen, do not allow of a very accurate means of comparison. The resistance of model (e) was 0.1365 pounds at the specified wind-tunnel air speed.TABLE OF BODY RESISTANCEThe speeds given in the above table are simply translational speeds, and are not corrected for slipstream velocity. With a slipstream of 25 per cent, increase the body resistance by 40 per cent. It would be safe to add an additional 10 per cent to make up for projecting fittings, baggy fabric, and scale variations.Since a body of approximately streamline form has a considerable percentage of skin friction, scale corrections for size and velocity are even of more importance than with wing sections. No wind-tunnel experiments can determine the resistance exactly because of the uncertainty of the scale factor. The resistance as given in the table is also affected by the proximity of the wing and tail surfaces, and by projections emanating from the motor compartment. It will be noted that the dirigible form B.F.-36 is markedly better than any of the others, being almost of perfect streamline form. The nearest approximation to the ideal form is N.P.L.-5, which has easy curves, low resistance, and is fairly symmetrical about the center line. Because of their small size, the pusher bodies or "nacelles" have a small total resistance, but the value of Kx is high.Chart Showing Forms of 7 Typical Aeroplane FuselageChart Showing Forms of 7 Typical Aeroplane FuselageProblem. Find the resistance of a Curtiss Tractor Type JN body with a breadth of 2' 6" and a depth of 3’ 3", the speed being 90 miles per hour. The slipstream is assumed to be 25 per cent, with an additional 10 per cent for added fabric loss, etc.Typical Stream Line Strut Construction.Typical Stream Line Strut Construction.Solution. The cross-sectional area = 2' 6" x 3’ 3" = A = 8.13 square feet. The velocity of translation is 90 M. P. H., or V² = 8100. The value of the resistance coefficient is taken from the table, Ko-0.00273. The total resistance R = KxAV² = 0.00273 x 8.13 x 8100 = 178.2 pounds. Since a slipstream of 25 per cent increases the resistance by 40 per cent, the resistance in the slipstream is 1782 x 1.4 = 249.48 pounds. The addition of the 10 per cent for extra friction makes the total resistance = 249.48 x 1.1 = 274.43 pounds. The resistance of this body, used with "twin" motors, would be 178.2 x 1.1 = 196.02, but as a tractor with the body in the slipstream, the resistance would be equal to 274.43 pounds as calculated above.

Effect of Resistance. Resistance to the forward motion of an aeroplane can be divided into two classes, (1) The resistance or drag due to the lift of the wings, and (2) The useless or "Parasitic" resistance due to the body, chassis and other structural parts of the machine. The total resistance is the sum of the wing drag and the parasitic resistance. Since every pound of resistance calls for a definite amount of power, it is of the greatest importance to reduce this loss to the lowest possible amount. The adoption of an efficient wing section means little if there is a high resistance body and a tangle of useless struts and wires exposed to the air stream. The resistance has a much greater effect on the power than the weight.

Weight and Resistance. We have seen that the average modern wing section will lift about 16 times the value of the horizontal drag, that is, an addition of 16 pounds will be equal to 1 pound of head resistance. If, by unnecessary resistance, we should increase the drag by 10 pounds, we might as well gain the benefit of 10 x 16 = 160 pounds of useful load. The higher the lift-drag efficiency of the wing, the greater will be the proportional loss by parasitic resistance.

Gliding Angle. The gliding angle, or the inclination of the path of descent when the machine is operating without power, is determined by the weight and the total head resistance. With a constant weight the angle is greatest when the resistance is highest. Aside from considerations of power, the gliding angle is of the greatest importance from the standpoint of safety. The less the resistance, and the flatter the angle of descent, the greater the landing radius.

Numerically this angle can be expressed by: Glide = W/R, where W = the weight of the aeroplane, and R = total resistance. Thus if the weight is 2500 pounds and the head resistance is 500 pounds, the rate of glide will be: 2500/500 = 5. This means that the machine will travel forward 5 feet for every foot that it falls vertically. If the resistance could be decreased to 100 pounds, the rate of glide would be extended to 2500/100 = 25, or the aeroplane would travel 25 feet horizontally for every foot of descent. This will give an idea as to the value of low resistance.

Resistance and Speed. The parasitic resistance of a body in uniform air varies as the square of the velocity at ordinary flight speeds. Comparing speeds of 40 and 100 miles per hour, the ratio will be as 40° is to 100° = 1600: 10,000 = 6.25, that is, the resistance at 100 miles per hour will be 6.25 times as great as at 40 miles per hour.

The above remarks apply only to bodies making constant angle with the air stream. Wings and lifting surfaces make varying angles at different speeds and hence do not show the same rate of increase. In carrying a constant load, the angle of the aeroplane wing is decreased as the speed increases and up to a certain point the resistance actually decreases with an increase in the speed. The wing resistance is greatest at extremely low speeds and at very high speeds. As the total resistance is made up of the sum of the wing and parasitic resistance at the different speeds, it does not vary according to any fixed law. The only true knowledge of the conditions existing through the range of flight speeds is obtained by drawing a curve in which the sums of the drag and head resistance are taken at intervals.

Resistance and Power. The power consumed in overcoming parasitic resistance increases at a higher rate than the resistance, or as the cube of the speed. Thus if the speed is increased from 40 to 100 miles per hour, the power will be increased 15.63 times. This can be shown by the following: Let V = velocity in miles per hour, H = Horsepower, K= Resistance coefficient of a body, A = Total area of presentation, and R = resistance in pounds. Then H = RV/375. Since R = KAV², then H = KAV² x V/375 = KAV³/375.

Resistance and Altitude. The resistance decreases with a reduction in the density of the air at constant speed. In practice, the resistance of an aeroplane is not in direct proportion to a decrease in the density as the speed must be increased at high altitudes in order to obtain the lift. The following example given by Capt. Green will show the actual relations.

Taking an altitude of 10,000 feet above sea level where the density is 0.74 of that at sea level, the resistance at equal speeds will be practically in proportion to the densities. In order to gain sustentation at the higher altitude, the speed must be increased, and hence the true resistance will be far from that calculated by the relative densities. Assume a sea level speed of 100 ft./sec., a weight of 3000 pounds, a lift-drag ratio of L/D = 15, and a body resistance of 40 pounds at sea level.

Because of the change in density at 10,000 feet, the flying speed will be increased from 100 feet per second to 350 feet per second in order to obtain sustentation. With sea level density this increase in speed (3.5 times) would increase the body resistance 3.5 x 3.5 = 12.25 times, making the total resistance 12.25 x 40 = 490 pounds. Since the density at the higher altitude is only 0.74 of that at sea level, this will be reduced by 0.26, or 0.26 x 490 = 364 pounds. Thus, the final practical result is that the sea level resistance of the body (40 pounds) is increased 9.1 times because of the speed increase necessary for sustentation. Since the wing angle and hence the liftdrag ratio would remain constant under both conditions, the wing drag would be constant at both altitudes, or 3000/15 = 200 pounds. The total sea level resistance at 100 feet per second is 200 + 40 = 240 pounds, while the total resistance at 10,000 feet becomes 364 + 200 = 564 pounds.

The speed varies as the square root of the change in density percentage. If V = velocity at sea level, v = velocity at a higher level, and d = percentage of the sea level density at the higher altitude, then v = V/√D. When the velocity at the high altitude is thus determined, the resistance can be easily obtained by the method given in Capt. Green's article. The following table gives the percentage of densities referred to sea level density.

Altitude Feet

Density Percent

Altitude Feet

Density Percent

Sea-level

1.00

7,500

0.78

1,000

.97

10,000

.74

2,000

.95

12,500

.66

3,000

.91

15,000

.61

5,000

.85

20,000

.52

If the velocity at sea level is 100 miles per hour, the velocity at 20,000 feet will be 100/0.72 = 139 miles per hour, where 0.72 is the square root of the density percentage, or the square root of 0.52 = .72 at 20,000 feet.

Total Parasitic Resistance. Aside from the drag of the wings, the resistance of the structural parts, body, tail and chassis depends upon the size and type of aeroplane. A speed scout has less resistance than a larger machine because of the small amount of exposed bracing, although the relative resistance of the body is much greater. The type of engine also has a great influence on the parasitic resistance. The following gives the approximate distribution of a modern fighting aeroplane:

Body

62 percent

Landing gear

16 "

Tail, fin, rudder

7 "

Struts, wires, etc.

15 "

The body resistance is by far the greatest item. A great part of the body resistance can be attributed to the motor cooling system, since in either case it is diverted from the true streamline form in order to accommodate the radiator, or the rotary motor cowl. The body resistance is also influenced by the necessity of accommodating a given cargo or passenger-carrying capacity, and by the distance of the tail surfaces from the wings. A body is not a streamline form when its length greatly exceeds 6 diameters.

Calculation of Total Resistance. The nearest approach that we can make to the actual head resistance by means of a formula is to adopt an expression in the form of R = KV² where K is a factor depending upon the size and type of machine. The true method would be to go over the planes and sum up the individual resistance of all the exposed parts. The parts lying in the propeller slip stream should be increased by the increased velocity of the slip stream. The parasitic resistance of biplanes weighing about 1800 pounds will average about, R = 0.036V² where V = velocity in miles per hour. Biplanes averaging 2500 pounds give R = 0.048V². Machines of the training or 2-seater type weigh from 1800 to 2500 pounds, and have an average head resistance distribution as follows:

Body, radiators, shields

35.5 percent.

Tail surface and bracing

14.9 "

Landing gear

17.2 "

Interplane struts, wires and fittings

23.6 "

Ailerons, aileron bracing, etc.

8.8 "

The averages in the above table differ greatly from the values given for the high speed fighting machine, principally because of the large control surfaces used in training machines, and the difference in the size of the motors.

With the wing drag being equal to D = Kx AV², and the total parasitic resistance equal to R = KV², the total resistance can be expressed by Rt = KxAV² + KV², where K = coefficient of parasitic resistance for different types and sizes of machines. The value of K for training machines will average 0.036, for machines weighing about 2500 pounds K = 0.048. Scouts and small machines will be safe at K = 0.028. The wing drag coefficient Kx varies with the angle of incidence and hence with the speed. For example, we will assume that the wing drag (Kx) of a scout biplane at 100 miles per hour is 0.00015, that the area is 200 square feet, and that the parasitic resistance coefficient is K = 0.028. The total resistance becomes: R = (0.00015 x 200 x 100 x 100) + 0.028 x 100 x 100 = 300 + 280 = 580 pounds. The formula in this case would be R = KxAV² + 0.028V².

Strut Resistance. The struts are of as nearly streamline form as possible. In practice the resistance must be compromised with strength, and for this reason the struts having the least resistance are not always applicable to the practical aeroplane. From the best results published by the N. P. L. the resistance was about 12.8 pounds per 100 feet strut at 60 miles per hour. The width of the strut is 1 inch. A rectangular strut under the same conditions gave a resistance of 104.4 pounds per 100 feet. A safe value would be 25 pounds per 100 feet at 60 miles per hour. If a wider strut is used, the resistance must be increased in proportion. With a greater speed, the resistance must be increased in proportion to the squares of the velocity. When the struts are inclined with the wind, the resistance is much decreased, and this is one advantage of a heavy stagger in a biplane.

The "Fineness ratio" or the ratio of the width to the depth of the section has a great effect on the resistance. With the depth equal to twice the width measured across the stream, a certain strut section gave a resistance of 24.8 pounds per 100 feet, while with a ratio of 3.5 the resistance was reduced to 11.4 pounds per hundred feet. Beyond this ratio the change is not as great, for with a ratio of 4.6 the resistance only dropped to 11.2 pounds.

Radiator Resistance. For the exact calculation of the radiator resistance it is first necessary to know the motor power and the fuel consumption since the radiator area, and hence the resistance, depends upon the size of the motor and the amount of heat transmitted to the jacket water. An aeronautic motor may be considered to lose as much through the water jackets as is developed in useful power, so that on this basis we should allow about 1.6 square feet of radiation surface per horsepower. This figure is arrived at by J. C. Hunsaker and assumes that the wind speed is 50 miles per hour (73 feet per second). The most severe cooling condition is met with in climbing at low speed, and it is here assumed that 50 miles per hour will represent the lowest speed that would be maintained for any length of time with the motor full out. For a racing aeroplane that will not climb for any length of time, one-half of the surface given above will be sufficient, and if the radiator is placed in the propeller slip stream it can be made relatively still smaller as the increased propeller slip at rapid rates of climb partially offsets the additional heating.

In the above calculations, Hunsaker does not take any particular type of radiator into consideration, merely assuming a smooth cooling surface. The Rome–Turney Company states that they allow 1.08 square feet of cooling surface per horsepower for honeycomb radiators, and 0.85 square feet for the helical tube type. The surface referred to means the actual surface measured all over the tubes and cells, and does not refer to the front area nor the exterior dimensions of the radiator. While a radiator may be made 25 percent smaller when placed in the slipstream, the resistance is increased by about 25 per cent, with a very small saving in weight, hence the total saving is small, if any. Side mounted radiators have a lower cooling effect per square foot than those placed in any other position, owing to the fact that the air must pass through a greater length of tube than where the broad side faces the wind.

In the radiator section tested by Hunsaker, there were about 64 square feet of cooling surface per square foot of front face area, but for absolute assurance on this point one should determine the ratio for the particular type of radiator that is to be used. The Auto Radiator Manufacturing Corporation, makers of the "Flexo" copper core radiators, have published some field tests made under practical conditions and for different types and methods of mounting. The four classes of radiators described are: (1) Front Type, in which the radiator is mounted in the end of the fuselage; (2) Side Type, mounted on the sides of the body; (3) Overhead Type, mounted above the fuselage and near the top plane; (4) Over-Engine Type, placed above and connected directly to the motor, as in the Standard H-3.

The following table gives the effectiveness of the different mountings in terms of the frontal area required per horsepower and the cooling surface, the area being in square inches (Front face area of radiator). Area in wind of type (3) is half the calculated frontal area since one core lies behind the other: Taking the value of the Rome–Turney honeycomb radiator as 60 square feet of cooling surface per horsepower, the frontal area per horsepower will be 0.0169 square feet, assuming that the radiator is approximately 6 inches thick. This amounts to 2.43 square inches of frontal area per horsepower.

Example. Find the approximate frontal area of a Rome-Turney type honeycomb radiator used with a motor giving 100 brake-horsepower. Find Resistance at 50 miles per hour (73 feet per second).

Class of Mounting

Square Inches Per H.P.

Cooling Surface Per H.P Square Inch

Front Type

4.00

117.00

Side Type

7.20

104.00

Overhead Type

2.70

112.00

Over-Engine Type

5.00

121.00

Solution. Area = A = 0.0169 HP = 0.0169 x 100 = 1.69 square feet. The honeycomb portion of surface for a square radiator of the above area will measure 16.2" x 16.2". Allowing a 1-inch water passage or frame all around the core, the side of the completed square radiator will measure 16.2" + 1" = 18.2". The diameter of a circular radiator core of the same 1.69 x 144 area will be 17.4 inches, since D = 1.69x144/0.7854. Adding the water passage, the overall diameter becomes 17.4 + 1 + 1 = 19.4 inches. The round honeycomb front radiator used on the 100 horsepower Curtiss Baby Scout measures 20 inches. Hunaker found the resistance of a honeycomb radiator to be R=0.000814 AV², there being 4 honeycomb cells per square inch. A = area of radiator in square feet, and V = velocity in feet per second. Adopting, for example, a speed of 73 feet per second, and an area equivalent to a 19.4-inch diameter circular radiator as above, the total resistance becomes:

R = 0.000814 AV² = 0.000814 x 3.1x (73 x 73) = 13.32 pounds, at 50 M. P. H. where A = 3.1 square feet.

Resistance of Chassis. Disc wheels (Enclosed spokes) have a resistance of about one-half that of open-wire wheels. The N. P. L. and Eiffel have agreed that the resistance of a wheel approximating 26" x 4" has a resistance of 1.7 pounds at 60 miles per hour (Disc type). For any other speed, the wheel resistance will be R = 1.7 V²/3600, where V = speed in miles per hour. We must also take into consideration the axle, chassis struts, wiring, shock absorbers, etc. The itemization of the chassis resistance, as given by the N.P.L. for the B.E.-2 biplane is as follows (60 miles per hour):

Wheels 2(a)1.75 pounds

3.5 pounds

Axle

2.0 "

Chassis struts and connections

1.1 "

Total chassis resistance(3)60 MPH.

6.6 pounds

At any other speed, the resistance for the complete chassis can be given by the formula R = 6.6V²/3600. This allowance will be ample, as the B. E.-2 is an old type and is equipped with skids.

Interplane Resistance. The interplane struts and wires are difficult to estimate by an approximate formula, the only exact way being to figure up each item separately from a preliminary drawing. The resistance varies with the form of the strut or wire section, the length, and the thickness. The fact that some of the struts lie in the propeller slipstream, and some outside of it, makes the calculation doubly difficult. The only recourse that we have at present is to analyze the conditions on the B. E.-2. With struts approximating true streamline form, a great percentage of the total resistance is skin friction, and as before explained, this item varies at a lesser rate than the square of the speed.

INTERPLANE RESISTANCE OF BIPLANE B.E-2 AT 60 M.P.H Table

According to a number of experiments on full size biplanes averaging 1900 pounds, it has been found that the interplane resistance (Struts, wires and fittings) amounts to about 24 per cent of the total parasitic head resistance of the entire machine, the drag of wings not being included. The maximum observed gave 29 per cent and the minimum 15 per cent. The resistance of the interplane bracing of speed scouts will be considerably less in proportion, as there are fewer exposed struts and cables on this type, the resistance probably averaging 15 per cent of the total head resistance. Based on these figures the resistance of the interplane bracing can be expressed by the following formula, in which I = resistance of interplane bracing in pounds, and V = translational speed in miles per hour:

I = 0.009 V² (For two-place biplanes weighing 1900 pounds).

I = 0.0054V² (For biplane speed scouts or racing type biplanes).

Strut Resistance. The above estimate includes wiring, strut fittings, etc., complete, and also takes the effect of the slipstream into consideration. A more accurate estimate can be made on the basis of strut length. To obtain this unit value we have recourse to the B. E.-2 tests. The translational speed in 60 miles per hour (88 feet per second) and the slipstream is taken at 25 feet per second. This gives a total velocity in the slipstream of 113 feet per second. The struts are 1% inches wide, and vary in length from 3’ 0" to 6’ 0". In the slipstream the increased velocity increases the resistance of the items by 64 per cent.

Total running length = 110’–0". Total resistance = 10.81 pounds. The resistance per foot = 10.81/110 = 0.099 pounds.

Resistance of Wire and Cable. In this estimate we will take the resistance given in the B. E.-2 tests, since values are given in the slipstream as well as for the outer portions. In the translational stream there is 240’ 0" of cable, 70’ 0" of No. 12 solid wire, and 52 turnbuckles, the total giving a resistance of 38.10 pounds. In the slipstream there is 50’ 0" of cable and 30’ 0" of solid wire with a resistance of 11.00 pounds. The total wire and cable resistance for the wings is therefore 49.10 pounds. The resistance of the wire and cable combined is 0.127 pounds per running foot.

Summary of Interplane Resistance. The total interplane resistance includes the struts, wires, cables and turnbuckles, a portion of which are in the slipstream. Since the total head resistance of the entire machine (B.E.-2) is 140 pounds at 60 M. P. H., and the interplane resistance = 10.81 + 49.10= 59.91 pounds, the relation of the interplane resistance to the total resistance is 43 per cent. This is much higher than the average (24 per cent), but the B.E.-2 is an old type of machine and the number of struts and wires were much greater than with modern aeroplanes.

Control Surface Resistance. The resistance of the control surfaces is a variable quantity, since so much depends upon the arrangement and form. Another variation occurring among machines of the same make and type is due to the various angles of the surfaces during flight, or at least during the time that they are used in correcting the attitude of the machine. With the elevator flaps or ailerons depressed to their fullest extent, the drag is many times that with the surfaces in "neutral," and as a general thing the controls are depressed at the time when the power demand is the greatest—that is, on landing, flying slow, or in "getting off."

Ailerons "in neutral" can be considered as being an integral part of the wings when they are hinged to the wing spar. In the older types of Curtiss machines the ailerons were hinged midway between the planes and the resistance was always in existence, whether the ailerons were in neutral or not. Wing warping, in general, can be assumed as in the case where the wings and ailerons are combined. With ailerons built into the wings, the resistance of the ailerons, and their wires and fittings, can be taken as being about 4 per cent of the total head resistance. With the aileron located between the two wings, the resistance may run as high as 20 per cent of the total.

Like the ailerons, the elevator surfaces and rudder are variable in attitude and therefore give a varying resistance. In neutral attitude the complete tail, consisting of the rudder, stabilizer, elevator, fin and bracing, will average about 15 per cent of the total resistance, it being understood that a non-lifting stabilizer is fitted. With lifting tails the resistance will be increased in proportion to the load carried by the stabilizer. In regard to the tail resistance it should be noted that these surfaces are in the slipstream and are calculated accordingly, although the velocity of the slipstream is somewhat reduced at the point where it encounters the tail surfaces. The total tail resistance of the B. E.-2 is given as 3.3 pounds.

Resistance of Seaplane Floats. The usual type of seaplane with double floats may be considered as having about 12 per cent higher resistance than a similar land machine. Some forms of floats have less resistance than others, owing to their better streamline form, but the above figure will be on the safe side for the average pontoon. Basing our formula on a 12 per cent increase on the total head resistance, the formula for the floats and bracing will become: Rt = 0.00436V² where R1 = resistance of floats and fittings.

Body Resistance. This item is probably the most difficult of any to compute, owing to the great variety of forms, the difference in the engine mounting, and the disposition of the fittings and connections. The resistance of the pilot's and passenger's heads, wind shields, and propeller arrangement all tend to increase the difficulty of obtaining a correct value. Aeroplanes with rotary air-cooled motors, or with large front radiators have a higher resistance than those arranged with other types of motors or radiator arrangements. Probably the item having the greatest influence on the resistance of the fuselage is the ratio of the length to the depth, or the "fineness ratio." In tractor monoplanes and biplanes, of the single propeller type, the body is in the slipstream, and compensation must be made for this factor.

If it were not for the motor and radiator, the tractor fuselage could be made in true dirigible streamline form, and would therefore present less resistance than the present forms of "practical" bodies. The necessity of placing the tail surfaces at a fixed distance from the wings also involves the use of a body that is longer in proportion than a true streamline form, and this factor alone introduces an excessive head resistance. The ideal ratio of depth to length would seem to range from 1 to 5.5 or 1 to 6. The fineness ratio of the average two-seat tractor is considerably greater than this, ranging from 1 to 7.5 or 8.5. A single-seat machine of the speed-scout type can be made much shorter and has more nearly the ideal proportions.

The only possible way of disposing of this problem is to compare the results of wind tunnel tests made on different types of bodies, and even with this data at hand a liberal allowance should be made because of the influence of the connections and other accessories. Eiffel, the N. P. L., and the Massachusetts Institute of Technology have made a number of experiments with scale models of existing aeroplane bodies. It is from these tests that we must estimate our body resistance, hence a table of the results is attached, the approximate outlines being shown by the figures.

As in calculating the resistance of other parts, the resistance of the body can be expressed by R = KxAV², where Kx = coefficient of the body form, A = Cross-sectional area of body in square feet (Area of presentation), and V = velocity in miles per hour. The area A is obtained by multiplying the body depth by the width. The "area of presentation" of a body 2’ 6" wide and 3’ 0" deep will be 2.5 x 3 = 7.5 square feet.

The experimental data does not give a very ready comparison between the different types, as the bodies not only vary in shape and size, but are also shown with different equipment. Some have tail planes and some have not; two are shown with the heads of the pilot and passenger projecting above the fuselage, while the remainder have either a simple cock-pit opening or are entirely closed. The presence of the propeller in two cases may have a great deal to do with raising the value of the experimental results. The propeller was stationary during the tests, but it was noted that the resistance was considerably less when the propeller was allowed to run as a windmill, driving the motor. This latter condition would correspond to the resistance in gliding with the motor cut off. In all cases, except the Deperdussin, the bodies are covered with fabric, and the sagging of the cloth in flight will probably result in higher resistance than would be indicated by the solid wood or metal model used in the tests. The pusher type bodies give less resistance than the tractors, but the additional resistance of the outriggers and tail bracing will probably bring the total far above the tractor body.

In the accompanying body chart are shown 7 representative bodies: (a) Deperdussin Monocoque Monoplane Body, a single-seater; (b) N. P. L.-5 Tractor Biplane Body, single-seater; (c) B. F.-36 Dirigible Form, without propeller or cock-pit openings; (d) B. E.-3. Two-Place Tractor Body, with passenger and pilot; (e) Curtiss JN Type Tractor Body, with passengers, chassis and tail; (f) Farman Pusher type, with motor, propeller and exposed passengers; (g) N. P. L. Pusher Body, bare. Body (a) was tested with a 1/5 scale model at a wind tunnel speed of 28 meters per second, the resistance of the model being 0.377 kilograms (0.83 pounds). Body (d) in model form was 1/16 scale and was tested at 20.5 miles per hour, at which speed the resistance was 0.0165 pounds. Model (e) was 1/12 scale and was tested at 30 miles per hour. These varying test speeds, it will be seen, do not allow of a very accurate means of comparison. The resistance of model (e) was 0.1365 pounds at the specified wind-tunnel air speed.

TABLE OF BODY RESISTANCE

The speeds given in the above table are simply translational speeds, and are not corrected for slipstream velocity. With a slipstream of 25 per cent, increase the body resistance by 40 per cent. It would be safe to add an additional 10 per cent to make up for projecting fittings, baggy fabric, and scale variations.

Since a body of approximately streamline form has a considerable percentage of skin friction, scale corrections for size and velocity are even of more importance than with wing sections. No wind-tunnel experiments can determine the resistance exactly because of the uncertainty of the scale factor. The resistance as given in the table is also affected by the proximity of the wing and tail surfaces, and by projections emanating from the motor compartment. It will be noted that the dirigible form B.F.-36 is markedly better than any of the others, being almost of perfect streamline form. The nearest approximation to the ideal form is N.P.L.-5, which has easy curves, low resistance, and is fairly symmetrical about the center line. Because of their small size, the pusher bodies or "nacelles" have a small total resistance, but the value of Kx is high.

Chart Showing Forms of 7 Typical Aeroplane FuselageChart Showing Forms of 7 Typical Aeroplane Fuselage

Chart Showing Forms of 7 Typical Aeroplane Fuselage

Problem. Find the resistance of a Curtiss Tractor Type JN body with a breadth of 2' 6" and a depth of 3’ 3", the speed being 90 miles per hour. The slipstream is assumed to be 25 per cent, with an additional 10 per cent for added fabric loss, etc.

Typical Stream Line Strut Construction.Typical Stream Line Strut Construction.

Typical Stream Line Strut Construction.

Solution. The cross-sectional area = 2' 6" x 3’ 3" = A = 8.13 square feet. The velocity of translation is 90 M. P. H., or V² = 8100. The value of the resistance coefficient is taken from the table, Ko-0.00273. The total resistance R = KxAV² = 0.00273 x 8.13 x 8100 = 178.2 pounds. Since a slipstream of 25 per cent increases the resistance by 40 per cent, the resistance in the slipstream is 1782 x 1.4 = 249.48 pounds. The addition of the 10 per cent for extra friction makes the total resistance = 249.48 x 1.1 = 274.43 pounds. The resistance of this body, used with "twin" motors, would be 178.2 x 1.1 = 196.02, but as a tractor with the body in the slipstream, the resistance would be equal to 274.43 pounds as calculated above.

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